/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 92 Solve for \(P: A=\frac{P t}{P+t}... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Solve for \(P: A=\frac{P t}{P+t}\)

Short Answer

Expert verified
\(P = \frac{-At}{A - t}\)

Step by step solution

01

Isolate the Fraction

In order to make it easier to solve for P, first multiply both sides by (P+t) to clear the fraction. This gives: A(P+t) = Pt.
02

Distribute A

Next, distribute A across the terms inside the parenthesis on the left side. This gives: AP + At = Pt.
03

Rearrange to Solve for P

Rearrange terms to isolate P on one side of the equation: AP - Pt = -At. This would allow us to factor out P from the left side of the equation.
04

Factor Out P

Now, take out P as a factor and rewrite the equation as: P(A - t) = -At.
05

Divide by (A-t)

Finally, to solve for P, divide both sides by (A - t) to get P on its own. This gives the final answer: \(P = \frac{-At}{A - t}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Fraction Clearing
Fraction clearing is an essential step in solving algebraic equations that include fractions. The goal is to remove the fraction by multiplying each side of the equation by the denominator. This simplifies the equation, making it easier to work with.

In the exercise, the equation was initially given as \(A = \frac{Pt}{P+t}\). To clear the fraction, we multiply both sides by \(P+t\) to remove the denominator. This results in \(A(P+t) = Pt\). The expression \(P+t\) is canceled out from the denominator, leaving a simple algebraic equation where both sides are free of fractions.
  • Clearing fractions helps to reduce complexity in solve-algebraic equations.
  • Ensures you have integer coefficients which are easier to manipulate.
Distributive Property
The distributive property is a crucial algebraic principle used to break down expressions. When you distribute a multiplication over an addition or subtraction inside parentheses, you ensure each term is multiplied by the factor outside the parentheses.

In our example, after clearing the fraction, we have \(A(P+t) = Pt\). Next, the distributive property allows us to expand this to \(AP + At = Pt\). This ensures each part of \(P+t\) is multiplied by \(A\).
  • Makes equations simpler by spreading out multiplication across terms.
  • Assists in unlocking the structure of an equation for further simplification.
Factoring
Factoring is a method used to simplify expressions or solve equations, where you express a term as a product of its factors. This technique often helps in revealing solutions and simplifying the problem.

Once we have the equation rearranged to \(AP - Pt = -At\), factoring allows us to "factor out" the common variable, \(P\), from the equation. This is done by recognizing \(P\) is common to both terms and rewriting the equation as \(P(A - t) = -At\). Now, \(P\) is isolated with a simple expression \((A-t)\), simplifying the path towards its solution.
  • Can turn complex expressions into more friendly forms.
  • Essential for isolating variables in algebra.
Isolation of Variable
The final goal in solving an equation is often to isolate the variable of interest. When the variable is isolated, it appears alone on one side of the equation, signifying it has been solved for.

The last steps of our solution involve isolating \(P\). After factoring, we have \(P(A - t) = -At\). To isolate \(P\), divide both sides by \(A-t\), giving \(P = \frac{-At}{A-t}\). Now, \(P\) stands alone on one side of the equation, providing the final solution.
  • Helps present the variable explicitly in terms of other constants or variables.
  • Results in a clear, straightforward solution suitable for interpretation.

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Most popular questions from this chapter

Rationalize the denominator: \(\frac{6}{\sqrt{3}-\sqrt{5}}\).

You buy a new car for 24,000 dollars. At the end of \(n\) years, the value of your car is given by the sequence $$a_{n}=24,000\left(\frac{3}{4}\right)^{n}, \quad n=1,2,3, \dots$$ Find \(a_{5}\) and write a sentence explaining what this value represents. Describe the \(n\) th term of the sequence in terms of the value of your car at the end of each year.

Will help you prepare for the material covered in the next section. Consider the sequence whose \(n\) th term is \(a_{n}=3 \cdot 5^{n}\). Find \(\frac{a_{2}}{a_{1}}, \frac{a_{3}}{a_{2}}, \frac{a_{4}}{a_{3}},\) and \(\frac{a_{5}}{a_{4}} .\) What do you observe?

Find a general term, \(a_{n},\) for each sequence. More than one answer may be possible. Evaluate without using a calculator: \(\frac{600 !}{599 !}\)

Use the formula for the sum of the first n terms of a geometric sequence to solve. A pendulum swings through an arc of 16 inches. On each successive swing, the length of the arc is \(96 \%\) of the previous length. \begin{aligned}&16, \quad\quad\quad\quad\quad 0.96(16), \quad(0.96)^{2}(16), \quad(0.96)^{3}(16), \ldots\\\&\begin{array}{|l|l|l|}\hline \begin{array}{l}\text { 1st swing } \\\\\end{array} & \begin{array}{l}\text{ 2nd swing } \\\\\end{array} & \begin{array}{l}\text { 3rd swing } \\\\\end{array} & \begin{array}{l}\text { 4th swing } \\\\\end{array} \\\\\hline\end{array}\end{aligned} After 10 swings, what is the total length of the distance the pendulum has swung? Round to the nearest hundredth of an inch.

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